Question

The file include/linux/typecheck.h of the Linux kernel 4.16 contains this code.

#define typecheck(type,x) \
({      type __dummy; \
        typeof(x) __dummy2; \
        (void)(&__dummy == &__dummy2); \
        1; \
}

which checks if x is the same type as the parameter type.

But I can’t understand the line:

 (void)(&__dummy == &__dummy2);

How does comparing the first address of both variables help?

Answer 1

Comparing pointers with incompatible types is a constraint violation and requires the compiler to issue a diagnostic. See 6.5.9 Equality operators:

Constraints

One of the following shall hold:

• both operands have arithmetic type;
• both operands are pointers to qualified or unqualified versions of compatible types;
• one operand is a pointer to an object or incomplete type and the other is a pointer to a qualified or unqualified version of void; or
• one operand is a pointer and the other is a null pointer constant.

and 5.1.1.3 Diagnostics:

A conforming implementation shall produce at least one diagnostic message (identified in an implementation-defined manner) if a preprocessing translation unit or translation unit contains a violation of any syntax rule or constraint, even if the behavior is also explicitly specified as undefined or implementation-defined. Diagnostic messages need not be produced in other circumstances.